Difference between revisions of "1950 AHSME Problems/Problem 3"
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<math> x^{2}-2x+\dfrac{5}{4}=0.</math> | <math> x^{2}-2x+\dfrac{5}{4}=0.</math> | ||
| − | Using Vieta's | + | The Vieta's formula states that in quadratic equation <math>ax^2+bx+c</math>, the sum of the roots of the equation is <math>-\frac{b}{a}</math>. |
| + | Using Vieta's formula, we find that the roots add to <math>2</math> or <math>\boxed{\textbf{(E)}\ \text{None of these}}</math>. | ||
==See Also== | ==See Also== | ||
| − | {{AHSME box|year=1950|num-b=2|num-a=4}} | + | {{AHSME 50p box|year=1950|num-b=2|num-a=4}} |
| + | |||
| + | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 01:05, 10 June 2020
Problem
The sum of the roots of the equation 
is equal to:
Solution
We can divide by 4 to get:
The Vieta's formula states that in quadratic equation 
, the sum of the roots of the equation is 
.
Using Vieta's formula, we find that the roots add to 
or 
.
See Also
| 1950 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
